NatAlgo total fees
    Updated 2022-06-26
    WITH tx as (
    SELECT date_trunc('day', b.block_timestamp) as date, SUM(fee) as total_algo_fee
    FROM algorand.transactions a
    INNER JOIN algorand.block b on a.block_id = b.block_id
    GROUP BY 1)

    SELECT date, total_algo_fee, SUM(total_algo_fee) OVER (ORDER BY date ASC) as fee_sum
    FROM tx
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